Complete topological vector space

The diagonal of X is the family[1] Δ_X  ≝  { (x, x) : x ∈ X }. For any N ⊆ X, the canonical entourage/vicinity around N is the set

Δ_X(N)  ≝  { (x, y) ∈ X × X  :  x - y ∈ N }  =  ∪y ∈ X [(y + N) × { y }]  =  Δ_X + (N × { 0 })

where if 0 ∈ N then Δ_X(N) contains the diagonal Δ_X  ≝  { (x, x) : x ∈ X  }  =  Δ_X({0}).

If N is symmetric (i.e. if - N = N), then Δ_X(N) is symmetric, which by definition means that (Δ_X(N))^op  ≝  { (y, x)  :  (x, y) ∈ Δ_X(N) }  =  Δ_X(N), and in addition, this symmetric set's composition with itself is:

Δ_X(N) ∘ Δ_X(N)  ≝  { (x, z) ∈ X × X  :  ∃ y ∈ X  such that  x, z ∈ y + N }  =  ∪y ∈ X [(y + N) × (y + N)]  =  Δ_X + (N × N).

If 𝒮 is any neighborhood basis at the origin in (X, τ) then the family of subsets of X × X :

ℬ_𝒮  ≝  { Δ_X(N)  :  N ∈ 𝒮 }

is a prefilter on X × X. If 𝒩_τ(0) is the neighborhood filter at the origin in (X, τ) then ℬ_𝒩τ(0) forms a base of entourages for a uniform structure on X that is considered canonical.[2] Explicitly, by definition, the canonical uniformity on X induced by (X, τ)[2] is the filter 𝒰_τ on X × X generated by the above prefilter:

𝒰_τ  ≝  ℬ_𝒩τ(0)^↑X×X  ≝  { S ⊆ X × X  :  N ∈ 𝒩_τ(0) and Δ_X(N) ⊆ S }

where ℬ_𝒩τ(0)^↑X×X denotes the upward closure of ℬ_τ in X × X.

If 𝒮 is any neighborhood basis at the origin in (X, τ) then the filter on X × X generated by the prefilter ℬ_𝒮 is equal to the canonical uniformity 𝒰_τ induced by (X, τ).

Uniform continuity

Let X and Y be TVSs, D ⊆ X, and f : D → Y be a map. Then f : D → Y is uniformly continuous if for every neighborhood U of the origin in X, there exists a neighborhood V of the origin in Y such that for all  x, y ∈ D,  if  y - x ∈ U  then  f (y) - f (x) ∈ V.

The general theory of uniform spaces has its own definition of a "Cauchy prefilter" and "Cauchy net". For the canonical uniformity on X, these definitions reduces down to the definition described below.

Suppose x_• = (x_i)_{i ∈ I} is a net in X and y_• = (y_i)_{j ∈ J} is a net in Y. The product I × J becomes a directed set by declaring (i, j) ≤ (i₂, j₂) if and only if i ≤ i₂ and j ≤ j₂. Then

x_• × y_•  ≝  (x_i, y_j)_{(i, j) ∈ I×J}

denotes the product net. If X = Y then the image of this net under the vector addition map X × X → X denotes the sum of these two nets:[3]

x_• + y_•  ≝  ( x_i + y_j )_{(i, j) ∈ I×J}

and similarly their difference is defined to be the image of the product net under the vector subtraction map:

x_• - y_•  ≝  ( x_i - y_j )_{(i, j) ∈ I×J}.

A net x_• = (x_i)_{i ∈ I} in a TVS X is called a Cauchy net[4] if

( x_i - x_j )_{(i, j) ∈ I×I}  →  0  in  X

or equivalently, if for every neighborhood N of 0 in X, there exists some i₀ ∈ I such that x_i - x_j ∈ N for all i, j ≥ i₀ with i, j ∈ I. A Cauchy sequence is a Cauchy net that is a sequence. It suffices to check any of these defining conditions for any given neighborhood basis of 0 in X.

A prefilter ℬ on an TVS (X, τ) called a Cauchy prefilter[5] if it satisfies any of the following equivalent conditions:

1. ℬ  -  ℬ  →  0  in X, where  ℬ  -  ℬ  ≝  { B  -  C  :  B, C ∈ ℬ } is a prefilter.
2. { B  -  B  :  B ∈ ℬ }  →  0  in X, where  { B  -  B  :  B ∈ ℬ } is a prefilter equivalent to ℬ  -  ℬ.
3. For every neighborhood N of 0 in X, ℬ contains some N-small set (i.e. there exists some B ∈ ℬ such that B - B ⊆ N).[6]
   • If B is a subset of a TVS X and N is a set containing 0, then B is N-small or small of order N[5] if B - B ⊆ N.
4. For every neighborhood N of 0 in X, there exists some B ∈ ℬ and some x ∈ X such that B ⊆ x + N.[5]

It suffices to check any of the above conditions for any given neighborhood basis of 0 in X.

If ℬ is a prefilter on a TVS X and x ∈ X then ℬ → x in X if and only if x ∈ cl ℬ and ℬ is Cauchy.[3]

For any S ⊆ X, a prefilter 𝒞 on S is necessarily a subset of ℘(S); that is, 𝒞 ⊆ ℘(S).

A subset S of a TVS (X, τ) is called complete if it satisfies any of the following equivalent conditions:

1. Every Cauchy prefilter 𝒞 ⊆ ℘(S) on S converges to at least one point of S.
   • If X is Hausdorff then every prefilter on S will converge to at most one point of X. But if X is not Hausdorff then a prefilter may converge to multiple points in X. The same is true for nets.
2. Every Cauchy net in S converges to at least one point of S.
3. S is a complete uniform space (under the point-set topology definition of "complete uniform space") when S is endowed with the uniformity induced on it by the canonical uniformity of X.

The subset S is called sequentially complete if every Cauchy sequence in S (or equivalently, every elementary Cauchy filter/prefilter on S) converges to at least one point of S.

Importantly, convergence outside of S is allowed: If X is not Hausdorff and if every Cauchy prefilter on S converges to some point of S, then S will be complete even if some or all Cauchy prefilters on S also converge to points(s) in X ∖ S. In short, there is no requirement that these Cauchy prefilters on S converge only to points in S. The same can be said of the convergence of Cauchy nets in S.

As a consequence, if a TVS X is not Hausdorff then every subset of the closure of { 0 } in X is complete because it is compact and every compact set is necessarily complete. In particular, if ∅ ≠ S ⊆ Cl_X { 0 } is a proper subset, such as S = { 0 } for example, then S would be complete even though every Cauchy net in S (and also every Cauchy prefilter on S) converges to every point in Cl_X { 0  }, including those points in Cl_X { 0 } that do not belong to S. This example also shows that complete subsets (and indeed, even compact subsets) of a non-Hausdorff TVS may fail to be closed. For example, if ∅ ≠ S ⊆ Cl_X { 0 } then S = Cl_X { 0 } if and only if S is closed in X.

A topological vector space (X, τ) is called complete if any of the following equivalent conditions are satisfied:

1. When endowed with its canonical uniformity, X becomes is a complete uniform space.
   • In the general theory of uniform spaces, a uniform space is called a complete uniform space if each Cauchy filter in X converges in (X, τ) to some point of X.
2. X is a complete subset of itself.
3. There exists a neighborhood of the origin in (X, τ) that is also a complete subset of X.[5]
   • This implies that every locally compact TVS is complete (even if the TVS is not Hausdorff).
4. Every Cauchy prefilter 𝒞 ⊆ ℘(X) on X converges in (X, τ) to at least one point of X.
   • If X is Hausdorff then every prefilter on X will converge to at most one point of X. But if X is not Hausdorff then a prefilter may converge to multiple points in X. The same is true for nets.
5. Every Cauchy filter on X converges in (X, τ) to at least one point of X.
6. Every Cauchy net in X converges in (X, τ) to at least one point of X.

where if in addition X is pseudometrizable or metrizable (e.g. a normed space) then this list can be extended to include:

7. (X, τ) is sequentially complete.

A topological vector space (X, τ) is sequentially complete if any of the following equivalent conditions are satisfied:

1. X is a sequentially complete subset of itself.
2. Every Cauchy sequence in X converges in (X, τ) to at least one point of X.
3. Every elementary Cauchy prefilter on X converges in (X, τ) to at least one point of X.
4. Every elementary Cauchy filter on X converges in (X, τ) to at least one point of X.

Let S ⊆ X, let Φ and Ψ be subsets of X × X, and let p ∈ X. Define[1]

Φ^op      ≝  { (y, x)  :  (x, y) ∈ Φ }

Φ ∘ Ψ  ≝  { (x, z)  :  ∃ y ∈ X  such that  (x, y) ∈ Ψ and (y, z) ∈ Φ }  =  ∪y ∈ X { (x, z)  :  (x, y) ∈ Ψ and (y, z) ∈ Φ }

S ⋅ Φ  ≝  { y ∈ X : Φ ∩ (S × { y }) ≠ ∅ }  =  Pr₂(Φ ∩ (S × X))

Φ ⋅ S  ≝  { x ∈ X : Φ ∩ ({ x } × S) ≠ ∅ }  =  Pr₁(Φ ∩ (X × S))  =  S ⋅ Φ^op

p ⋅ Φ  ≝  { p } ⋅ Φ = { y ∈ X  :  (p, y) ∈ Φ }

Φ ⋅ p  ≝  Φ ⋅ { p } = { x ∈ X  :  (x, p) ∈ Φ }  =  p ⋅ Φ^op

where Φ ⋅ S (resp. S ⋅ Φ) is called the set of left (resp. right) Φ-relatives of (points in) S. The maps Pr₁ and Pr₂ are the canonical projections onto the first and second coordinates, respectively. Given Φ ⊆ X × X, Φ is symmetric if Φ = Φ^op while a subset S ⊆ X is called Φ-small if S × S ⊆ Φ. Two points x and y are Φ-close if (x, y) ∈ Φ.

If Φ, Ψ, Ω ⊆ X × X and R, S ⊆ X then:

• Φ ⊆ Ψ if and only if Φ ^op ⊆ Ψ ^op. Moreover, (Φ ∘ Ψ)^op  =  Ψ ^op ∘ Φ ^op and Φ ∩ Ψ is symmetric.
• If Φ ⊆ Φ₂ and Ψ ⊆ Ψ₂ then Φ ∘ Ψ  ⊆  Φ₂ ∘ Ψ₂.
• Associativity: Φ ∘ (Ψ ∘ Ω)  =  (Φ ∘ Ψ) ∘ Ω.
• (Φ ∘ Ψ) ⋅ S = Φ ⋅ (Ψ ⋅ S).

A non-empty family ℬ ⊆ ℘(X × X) is called a base or fundamental system of entourages if ℬ is a prefilter on X × X satisfying all of the following conditions:

1. Every set in ℬ contains the diagonal of X as a subset; that is, Δ_X  ≝  { (x, x) : x ∈ X } ⊆ Φ for every Φ ∈ ℬ. Said differently, the prefilter ℬ is fixed on Δ_X.
2. For every Ω ∈ ℬ there exists some Φ ∈ ℬ such that Φ ∘ Φ ⊆ Ω.
3. For every Ω ∈ ℬ there exists some Φ ∈ ℬ such that Φ ⊆ Ω^op.

A uniformity or uniform structure on X is a filter 𝒰 on X × X that is generated by some base of entourages ℬ, in which case we say that ℬ is a base of entourages for 𝒰.

For a commutative additive group X, a fundamental system of entourages ℬ is called translation-invariant[7] if for every Φ ∈ ℬ, (x, y) ∈ Φ if and only if (x + z, y + z) ∈ Φ for all x, y, z ∈ X. A uniformity ℬ is called translation-invariant[7] if it has a base of entourages that is translation-invariant. The same canonical uniformity would result by using a neighborhood basis of the origin rather the filter of all neighborhoods of the origin. The canonical uniformity on any TVS is translation-invariant.[7]

Let ℬ ⊆ ℘(X × X) be a base of entourages on X. For every S ⊆ X and p ∈ X, the neighborhood prefilter on S (resp. at p) is the set

ℬ ⋅ S  ≝  { Φ ⋅ S  :  Φ ∈ ℬ }       and       ℬ ⋅ p  ≝  ℬ ⋅ { p }  =  { Φ ⋅ p  :  Φ ∈ ℬ }

and the filter on X that it generates is known as the neighborhood filter of S (resp. of p). The assignment

x ∈ X ↦ ℬ ⋅ x ≝ { Φ ⋅ x  :  Φ ∈ ℬ }

of points to prefilters generates a topology on X called the topology induced by ℬ. A subset U ⊆ X is open in this topology if and only if any of the following equivalent conditions hold:

1. For every u ∈ U there exists some N ∈ ℬ ⋅ u such that N ⊆ U.
2. For every u ∈ U there exists some Φ ∈ ℬ such that Φ ⋅ u = { x ∈ X  :  (x, u) ∈ Φ } ⊆ U.

The closure of a subset S ⊆ X in this topology is:

 Cl_X S  =  ∩Φ ∈ ℬ (Φ ⋅ S)  =  ∩Φ ∈ ℬ (S ⋅ Φ).

Cauchy prefilters and complete uniformities

A prefilter ℱ ⊆ ℘(X) on a uniform space X with uniformity 𝒰 is called a Cauchy prefilter if for every entourages N ∈ 𝒰, there exists some F ∈ ℱ such that F × F ⊆ N.

A uniform space (X, 𝒰) is called complete (resp. sequentially complete) if every Cauchy prefilter (resp. every elementary Cauchy prefilter) on X converges to at least one point of X when X is endowed with the topology induced by 𝒰.

If (X, τ) is a TVS then for any S ⊆ X and x ∈ X,

Δ_X(N) ⋅ S  =  S + N      and      Δ_X(N) ⋅ x  =  x + N.

The topology induced on X by the canonical uniformity is the same as the topology that X started with (i.e. it is τ).

We review the basic notions related to the general theory of complete pseudometric spaces. Recall that every metric is a pseudometric and that a pseudometric p is a metric if and only if p(x, y) = 0 implies x = y. Thus every metric space is a pseudometric space and a pseudometric space (X, p) is a metric space if and only if p is a metric.

If S is a subset of a pseudometric space (X, d) then the diameter of S is defined to be

diam(S) ≝ sup { d (s, t) : s, t ∈ S  }.

A prefilter ℬ on a pseudometric space (X, d) is called a d-Cauchy prefilter or simply a Cauchy prefilter if for each real r > 0, there is some B ∈ ℬ such that the diameter of B is less than r.

Suppose (X, d) is a pseudometric space. A net x_• = (x_i)_{i ∈ I} in X is called a d-Cauchy net or simply a Cauchy net if Tails (x_•) is a Cauchy prefilter, which happens if and only if

for every r > 0 there is some i ∈ I such that if j, k ∈ I with j  ≥  i and k  ≥  i then d (x_j, x_jn)  <  r.

A Cauchy sequence is a sequence that is also a Cauchy net.[note 2]

Every pseudometric p on a set X induces the usual canonical topology on X, which we'll denote by τ_p; it also induces a canonical uniformity on X, which we'll denote by 𝒰_p. The topology on X induced by the uniformity 𝒰_p is equal to τ_p. A net x_• = (x_i)_{i ∈ I} in X is Cauchy with respect to p if and only if it is Cauchy with respect to the uniformity 𝒰_p. The pseudometric space (X, p) is a complete (resp. a sequentially complete) pseudometric space if and only if (X, 𝒰_p) is a complete (resp. a sequentially complete) uniform space. Moreover, the pseudometric space (X, p) (resp. the uniform space (X, 𝒰_p)) is complete if and only if it is sequentially complete.

A pseudometric space (X, d) (e.g. a metric space) is called complete and d is called a complete pseudometric if any of the following equivalent conditions hold:

1. Every Cauchy prefilter on X converges to at least one point of X.
2. The above statement but with the word "prefilter" replaced by "filter."
3. Every Cauchy net in X converges to at least one point of X.
4. Every Cauchy sequence in X converges to at least one point of X.
   • Thus to prove that (X, d) is complete, it suffices to only consider Cauchy sequences in X (and it is not necessary to consider the more general Cauchy nets).
5. The canonical uniformity on X induced by the pseudometric d is a complete uniformity.

• Note that if d is a metric on X then any such limit point is necessarily unique and the same is true for limits of Cauchy prefilters on X.

And if addition d is a metric then we may add to this list:

6. Every decreasing sequence of closed balls whose diameters shrink to 0 has non-empty intersection.[8]

Every F-space, and thus also every Fréchet space, Banach space, and Hilbert space is a complete TVS. Note that every F-space is a Baire space but there are normed spaces that are Baire but not Banach.[9]

A pseudometric d on a vector space X is said to be translation invariant if d(x,  y) = d(x + z,  y + z) for all vectors x, y, z ∈ X.

Suppose (X, τ) is pseudometrizable TVS (e.g. a metrizable TVS) and that p is any pseudometric on X such that the topology on X induced by p is equal to τ. If p is translation-invariant, then (X, τ) is a complete TVS if and only if (X, p) is a complete pseudometric space.[10] If p is not translation-invariant, then may be possible for (X, τ) to be a complete TVS but (X, p) to not be a complete pseudometric space[10] (see this footnote[note 3] for an example).[10]

Two norms on a vector space are called equivalent if and only if they induce the same topology.[13] If p and q are two equivalent norms on a vector space X then the normed space (X, p) is a Banach space if and only if (X, q) is a Banach space. See this footnote for an example of a continuous norm on a Banach space that is not is not equivalent to that Banach space's given norm.[note 5][13] All norms on a finite-dimensional vector space are equivalent and every finite-dimensional normed space is a Banach space.[14]

As the example below shows, regardless of whether or not a space is Hausdorff or already complete, every topological vector space (TVS) has infinitely many non-isomorphic completions.[16]

However, every Hausdorff TVS has a Hausdorff completion that is unique up to TVS-isomorphism.[16] But nevertheless, every Hausdorff TVS still has infinitely many non-isomorphic non-Hausdorff completions.

Example (Non-uniqueness of completions):[15] Let C denote any complete TVS and let I denote any TVS endowed with the indiscrete topology, which recall makes I into a complete TVS. Since both I and C are complete TVSs, so is their product I × C. If U and V are non-empty open subsets of I and C, respectively, then U = I and (U × V) ∩ ({ 0 } × C) = { 0 } × V ≠ ∅, which shows that { 0 } × C is a dense subspace of I × C. Thus by definition of "completion," I × C is a completion of { 0 } × C  (it doesn't matter that { 0 } × C is already complete). So by identifying { 0 } × C with C, if X ⊆ C is a dense vector subspace of C, then X has both C and I × C as completions.

Every Hausdorff TVS has a Hausdorff completion that is unique up to TVS-isomorphism.[16] But nevertheless, as shown above, every Hausdorff TVS still has infinitely many non-isomorphic non-Hausdorff completions.

Existence of Hausdorff completions

A Cauchy filter ℬ on a TVS X is called a minimal Cauchy filter[17] if there does not exist a Cauchy filter on X that is strictly coarser than ℬ (i.e. "strictly coarser than ℬ" means contained as a proper subset of ℬ).

If ℬ is a Cauchy filter on X then the filter generated by the following prefilter:

{ B + N  :  B ∈ ℬ and N is a neighborhood of 0 in X }

is the unique minimal Cauchy filter on X that is contained as a subset of ℬ.[17] In particular, for any x ∈ X, the neighborhood filter at x is a minimal Cauchy filter.

Let 𝕄 be the set of all minimal Cauchy filters on X and let E : X → 𝕄 be the map defined by sending x ∈ X to the neighborhood filter of x in X. We endow 𝕄 with the following vector space structure. Given ℬ, 𝒞 ∈ 𝕄 and a scalar s, let ℬ + 𝒞 (resp. s ℬ) denote the unique minimal Cauchy filter contained in the filter generated by { B + C  :  B ∈ ℬ and C ∈ 𝒞 } (resp. { sB  :  B ∈ ℬ  }).

For every balanced neighborhood N of 0 in X, let

𝕌(N)  ≝  { ℬ ∈ 𝕄  :  ∃ B ∈ ℬ and ∃ a neighborhood V of 0 in X such that B + V ⊆ N  }.

If X is Hausdorff then the collection of all sets 𝕌(N), as N ranges over all balanced neighborhoods of 0 in X, forms a vector topology on 𝕄 making 𝕄 into a complete Hausdorff TVS. Moreover, the map E : X → 𝕄 is a TVS-embedding onto a dense vector subspace of 𝕄.[17]

If X is a metrizable TVS then a Hausdorff completion of X can be constructed using equivalence classes of Cauchy sequences instead of minimal Cauchy filters.

We now show how every non-Hausdorff TVS X can be TVS-embedded onto a dense vector subspace of a complete TVS. The proof that every Hausdorff TVS has a Hausdorff completion is widely available so we use its conclusion to prove that every non-Hausdorff TVS also has a completion. These details are sometimes useful for extending results from Hausdorff TVSs to non-Hausdorff TVSs.

Let I = cl { 0 } denote the closure of the origin in X, where I is endowed with its subspace topology induced by X (so that I has the indiscrete topology). Since I has the trivial topology, it is easily shown that every vector subspace of X that is an algebraic complement of I in X is necessarily a topological complement of I in X.[18][19] Let H denote any topological complement of I in X, which is necessarily a Hausdorff TVS (since it is TVS-isomorphic to the quotient TVS X / I [note 6]). Since X is the topological direct sum of I and H (which means that X = I ⊕ H in the category of TVSs), the canonical map

I × H  →  I ⊕ H = X,     given by     (x, y)  ↦  x + y

is a TVS-isomorphism.[19] Let A  :  X = I ⊕ H  →  I × H denote the inverse of this canonical map. (As a side note, it follows that every open and every closed subset U of X satisfies U = I + U.[proof 1])

The Hausdorff TVS H can be TVS-embedded, say via the map In_H  :  H  →  C, onto a dense vector subspace of its completion C. Since I and C are complete, so is their product I × C. Let Id_I : I → I denote the identity map and observe that the product map Id_I × In_H  :  I × H  →  I × C is a TVS-embedding whose image is dense in I × C. Define the map[note 7]

B  :  X = I ⊕ H  →  I × C     by     B  ≝  (Id_I × In)  ∘  A

which is a TVS-embedding of X = I ⊕ H onto a dense vector subspace of the complete TVS I × C. Moreover, observe that the closure of the origin in I × C is equal to I × { 0  }, and that I × { 0 } and { 0 } × C are topological complements in I × C.

To summarize,[19] given any algebraic (and thus topological) complement H of I ≝ cl { 0 } in X and given any completion C of the Hausdorff TVS H such that H ⊆ C, then the natural inclusion[20]

In_H  :  X = I ⊕ H  →  I ⊕ C

is a well-defined TVS-embedding of X onto a dense vector subspace of the complete TVS  I ⊕ C where moreover, we have

X  =  I ⊕ H  ⊆  I ⊕ C  ≅  I × C.

Said differently, if C is a completion of a TVS X with X ⊆ C and if 𝒩 is a neighborhood base of the origin in X, then the family of sets

{ Cl_C N  :  N ∈ 𝒩 }

is a neighborhood basis at the origin in C.[3]

If a TVS X has any of the following properties then so does its completion:

• Hausdorff
• Locally convex
• Pseudometrizable[16]
• Metrizable[16]
• Seminormable
• Normable
  • Moreover, if (X, p) is a normed space, then the completion can be chosen to be a Banach space (B, q) such that the TVS-embedding of (X, p) into (B, q) is an isometry.
• Hausdorff pre-Hilbert. That is, a TVS induced by an inner product.[23]
  • Every inner product space (H, ⟨·, ·⟩) has a completion (H, ⟨·, ·⟩_H) that is a Hilbert space, where the inner product ⟨·, ·⟩_H is the unique continuous extension to H of the original inner product ⟨·, ·⟩. The norm induced by (H, ⟨·, ·⟩_H) is also the unique continuous extension to H of the norm induced by ⟨·, ·⟩.[23][21]
• Nuclear[22]
• Barrelled[24]
• Mackey[25]
• DF-space[26]

Other preserved properties

If X is a Hausdorff TVS, then the continuous dual space of X is identical to the continuous dual space of the completion of X.[27] The completion of a locally convex bornological space is a barrelled space.[24] If X and Y are DF-spaces then the projective tensor product, as well as its completion, of these spaces is a DF-space.[28]

The completion of the projective tensor product of two nuclear spaces is nuclear.[22] The completion of a nuclear space is TVS-isomorphic with a projective limit of Hilbert spaces.[22]

If f : X → Y is a nuclear linear operator between two locally convex spaces and if C be a completion of X then f has a unique continuous linear extension to a nuclear linear operator F : C → Y.[22]

Let X and Y be two Hausdorff TVSs with Y complete. Let C be a completion of X. Let L(X; Y) denote the vector space of continuous linear operators and let I : L(X; Y) → L(C; Y) denote the map that sends every f ∈ L(X; Y) to its unique continuous linear extension on C. Then I : L(X; Y) → L(C; Y) is a (surjective) vector space isomorphism. Moreover, I : L(X; Y) → L(C; Y) maps families of equicontinuous subsets onto each other. Suppose that L(X; Y) is endowed with a 𝒢-topology and that ℋ denotes the closures in C of sets in 𝒢. Then the map I : L_𝒢(X; Y) → L_ℋ(C; Y) is also a TVS-isomorphism.[22]

Every TVS has a completion and every Hausdorff TVS has a Hausdorff completion.[33] Every complete TVS is quasi-complete space and sequentially complete.[34] However, the converses of the above implications are generally false.[34] There exists a sequentially complete locally convex TVS that is not quasi-complete.[26]

If a TVS has a complete neighborhood of the origin then it is complete.[35] Every complete pseudometrizable TVS is a barrelled space and a Baire space (and thus non-meager).[36] The dimension of a complete metrizable TVS is either finite or uncountable.[19]

Any neighborhood basis of any point in a TVS is a Cauchy prefilter.

Every convergent net (resp. prefilter) in a TVS is necessarily a Cauchy net (resp. a Cauchy prefilter).[5] Any prefilter that is subordinate to (i.e. finer than) a Cauchy prefilter is necessarily also a Cauchy prefilter[5] and any prefilter finer than a Cauchy prefilter is also a Cauchy prefilter. The filter associated with a sequence in a TVS is Cauchy if and only if the sequence is a Cauchy sequence. Every convergent prefilter is a Cauchy prefilter.

If X is a TVS and if x ∈ X is a cluster point of a Cauchy net (resp. Cauchy prefilter), then that Cauchy net (resp. that Cauchy prefilter) converges to x in X.[3] If a Cauchy filter in a TVS has an accumulation point x then it converges to x.

Uniformly continuous maps send Cauchy nets to Cauchy nets.[3] A Cauchy sequence in a Hausdorff TVS X is not necessarily relatively compact (i.e. its closure in X is not necessarily compact).

Suppose that X_• = (X_i)_{i ∈ I} is a family of TVSs and that X denotes the product of these TVSs. Suppose that for every index i, ℬ_i is a prefilter on X_i. Then the product of this family of prefilters is a Cauchy filter on X if and only if each ℬ_i is a Cauchy filter on X_i.[17]

If f : X → Y is an injective topological homomorphism from a complete TVS into a Hausdorff TVS then the range of f (i.e. f (X)) is a closed subspace of Y.[31] If f : X → Y is a topological homomorphism from a complete metrizable TVS into a Hausdorff TVS then the range of f is a closed subspace of Y.[31] If f : X → Y is a uniformly continuous map between two Hausdorff TVSs then the image under f of a totally bounded subset of X is a totally bounded subset of Y.[37]

Uniformly continuous extensions

Suppose that f : D → Y is a uniformly continuous map from a dense subset D of a TVS X into a complete Hausdorff TVS Y. Then f has a unique uniformly continuous extension to all of X.[3] If in addition f is a homomorphism then its unique uniformly continuous extension is also a homomorphism.[3] This remains true if "TVS" is replaced by "commutative topological group."[3] The map f is not required to be a linear map and that D is not required to be a vector subspace of X.

Uniformly continuous linear extensions

Suppose f : X → Y be a continuous linear operator between two Hausdorff TVSs. If M is a dense vector subspace of X and if the restriction f |_M : M → Y to M is a topological homomorphism then f : X → Y is also a topological homomorphism.[38] So if C and D are Hausdorff completions of X and Y, respectively, and if f : X → Y is a topological homomorphism, then f 's unique continuous linear extension F : C → D is a topological homomorphism. (Note that it's possible for f : X → Y to be surjective but for F : C → D to not be injective.)[38]

Suppose X and Y are Hausdorff TVSs, M is a dense vector subspace of X, and N is a dense vector subspaces of Y. If M are and N are topologically isomorphic additive subgroups via a topological homomorphism f then the same is true of X and Y via the unique uniformly continuous extension of f (which is also a homeomorphism).[39]

Complete subsets

Every complete subset of a TVS is sequentially complete. A complete subset of a Hausdorff TVS X is a closed subset of X.[3][35]

Every compact subset of a TVS is complete (even if the TVS is not Hausdorff or not complete).[3][35] Closed subsets of a complete TVS are complete; however, if a TVS X is not complete then X is a closed subset of X that is not complete. The empty set is complete subset of every TVS. If C is a complete subset of a TVS (the TVS is not necessarily Hausdorff or complete) then any subset of C that is closed in C is complete.[35]

Topological complements

If X is a non-normable Fréchet space on which there exists a continuous norm then X contains a closed vector subspace that has no topological complement.[26] If X is a complete TVS and M is a closed vector subspace of X such that X / M is not complete, then H does not have a topological complement in X.[26]

Subsets of completions

Let M be a separable locally convex metrizable topological vector space and let C be its completion. If S is a bounded subset of C then there exists a bounded subset R of X such that S ⊆ cl_C R.[26]

Relation to compact subsets

A subset of a TVS (not assumed to be Hausdorff or complete) is compact if and only if it is complete and totally bounded.[40][proof 2] Thus a closed and totally bounded subset of a complete TVS is compact.[41][3]

Every complete totally bounded set is relatively compact.[3] If X is any TVS then the quotient map q : X → X / cl_X({ 0 }) is a closed map[42] and thus S + cl_X ({ 0 })  ⊆  Cl_X S A subset S of a TVS X is totally bounded if and only if its image under the canonical quotient map q : X → X / cl_X({ 0 }) is totally bounded.[19] Thus S is totally bounded if and only if S + cl_X ({ 0 }) is totally bounded. In any TVS, the closure of a totally bounded subset is again totally bounded.[3] In a locally convex space, the convex hull and the disked hull of a totally bounded set is totally bounded.[33] If S is a subset of a TVS X such that every sequence in S has a cluster point in S then S is totally bounded.[19] A subset S of a Hausdorff TVS X is totally bounded if and only if every ultrafilter on S is Cauchy, which happens if and only if it is pre-compact (i.e. its closure in the completion of X is compact).[37]

If S ⊆ X is compact, then cl_X S  = S + cl_X { 0 } and this set is compact. Thus the closure of a compact set is compact[note 8] (i.e. all compact sets are relatively compact).[43] Thus the closure of a compact set is compact. Every relatively compact subset of a Hausdorff TVS is totally bounded.[37]

In a complete locally convex space, the convex hull and the disked hull of a compact set are both compact.[33] More generally, if K is a compact subset of a locally convex space, then the convex hull co K (resp. the disked hull cobal K) is compact if and only if it is complete.[33] Every subset S of cl({ 0 }) is compact and thus complete.[proof 3] In particular, if X is not Hausdorff then there exist compact complete sets that are not closed.[3]